Unformatted text preview: AP Physics C: Mechanics
2001 Scoring Guidelines The materials included in these files are intended for noncommercial use by AP
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Copyright © 2001 by College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, and the acorn logo are registered
trademarks of the College Entrance Examination Board. AP® PHYSICS C: MECHANICS
2001 SCORING GUIDELINES General Notes about 2001 AP Physics Solutions
1. The solutions contain the most common method(s) of solving the freeresponse questions, and the
allocation of points for these solutions. Other methods of solution also receive appropriate credit for
correct work.
2. Generally, double penalty for errors is avoided. For example, if an incorrect answer to part (a) is
correctly substituted into an otherwise correct solution to part (b), full credit will usually be awarded.
3. An exception to this may be cases when the numerical answer to a later part should be easily
recognized as wrong, e.g., a speed faster than the speed of light in vacuum.
4. Implicit statements of concepts normally receive credit. For example, if use of the equation
expressing a particular concept is worth one point, and a solution contains the application of the
equation to the problem but does not separately list the basic equation, the point is still awarded. Copyright © 2001 by College Entrance Examination Board. All rights reserved.
Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 2 AP® PHYSICS C: MECHANICS
2001 SCORING GUIDELINES
Question 1
15 points total
1. (a) 6 points Distribution
of Points The average acceleration is the change in velocity divided by the time interval
For correct subtraction to find the time interval ∆t = t f − ti = 0.37 − 0.33 = 0.04 s 1 point From graph: υi = 0.22 m s
For getting υi in the range 0.2 < υi ≤ 0.25 m s 1 point From graph: υ f = −0.18 m s
For getting υ f in the range −0.15 ≥ υ f > −0.20 m s 1 point For getting ∆υ consistent with the student’s values of υi and υ f , including subtracting in 1 point the correct direction ∆υ = υ f − υi = −0.18 m s − 0.22 m s = −0.40 m s a= ∆υ −0.4 m s
=
∆t
0.04 s For correct substitution of values in the above equation a = −10 m s 1 point 2 For showing deceleration (e.g., with a minus sign)
1 point Note: There were three alternate methods for solving parts (b) and (c) that could receive full credit.
Method 1.
1. (b) 3 points For any indication of the concept of finding the area under the curve in the second graph 1 point ∆p = ò F dt or ∆ p = the area under the F vs. t curve
∆p = 0.6 N • s or ∆p = 0.6 kg • m
s For correct numerical value of 0.6
For correct units
1. (c) 1 point
1 point 2 points For any statement of the correct equation for the change in momentum 1 point For correct substitution of values consistent with those obtained above 1 point ∆p = m∆υ
m= ∆p 0.6 N s
=
= 1.5 kg
∆υ 0.4 m s
• Copyright © 2001 by College Entrance Examination Board. All rights reserved.
Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 3 AP® PHYSICS C: MECHANICS
2001 SCORING GUIDELINES
Question 1 (cont.)
Method 2.
1. (b) 3 points Distribution
of Points Expressing the change in momentum in terms of the average force: ∆p = F ∆t
For some method of using the graph to find the average force for the four nonzero
intervals such as indicating that the area is equivalent to 6 boxes each with a height 1 point of 10 N, so that F = 60 4 = 15 N ∆p = (15 N )( 0.04 s ) =0.6 N • s or ∆p = 0.6 kg • m
s For correct numerical value of 0.6
For correct units
1. (c) 1 point
1 point 2 points Expressing the average force in terms of the average acceleration: F = ma For correct equation ( F = ma also accepted)
For correct substitution of values consistent with those obtained above m= 1 point
1 point 15 N
F
=
= 1.5 kg
a 10 m s 2 Method 3. Student solved part (c) first and went back to part (b)
1. (c) 3 points For some method of using the graph to find the average force for the four nonzero
intervals such as indicating that the area is equivalent to 6 boxes each with a height
of 10 N, so that F = 60 4 = 15 N
For a correct expression for Newton’s second law F = ma For correct substitution of values consistent with those obtained above m= 1 point 1 point
1 point 15 N
F
=
= 1.5 kg
a 10 m s 2 Copyright © 2001 by College Entrance Examination Board. All rights reserved.
Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 4 AP® PHYSICS C: MECHANICS
2001 SCORING GUIDELINES
Question 1 (cont.)
Method 3. (continued)
1. (b) 2 points Distribution
of Points ∆p = m∆υ = (1.5 kg ) ( 0.4 m s ) =0.6 N • s or ∆p = 0.6 kg • m
s For correct substitution of values consistent with those obtained above
For correct units
1. (d) 1 point
1 point 4 points For a correct statement of energy change 1 point ∆E = E f − Ei
For a kinetic energy equation 1 point 1
E = mυ 2
2
For correct substitution of values consistent with those obtained above including the
squared velocities 1
1
2
2
(1.5 kg ) ( 0.22 m s ) − (1.5 kg ) ( −0.18 m s )
2
2
∆E = 0.012 J 1 point ∆E = For correct units 1 point Copyright © 2001 by College Entrance Examination Board. All rights reserved.
Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 5 AP® PHYSICS C: MECHANICS
2001 SCORING GUIDELINES
Question 2
15 points total
2. (a) i. 3 points Distribution
of Points There were two methods generally used to solve this part.
Method 1. F = mac
GM J m
F=
R2
For a statement of at least one of Newton’s laws above
Equating the two equations above and substituting expression for centripetal force: mac = 1 point GM J m
R2 For substituting the expression for centripetal force 1 point mυ
GM J m
=
R
R2
2 For a solution for υ that follows algebraically from previous work 1 point GM J
R υ= Method 2. a= GM J
R2 or g= GM J
R2 For statement of either of the above, which are derived from Newton’s laws
For a correct statement of centripetal acceleration ac = 1 point
1 point υ2
R Equating the two expressions above for ac and solving for υ : υ= GM J
R For a solution for υ that follows algebraically from previous work.
Two points were awarded for an approach that started with K = 1 point 1
GM J m
mυ 2 =
and
2
2R solved for υ as long as there was no sign error in the equation and there were no incorrect
statements regarding energy prior to the equation.
Copyright © 2001 by College Entrance Examination Board. All rights reserved.
Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 6 AP® PHYSICS C: MECHANICS
2001 SCORING GUIDELINES
Question 2 (cont.)
2. (a) ii. 3 points Distribution
of Points There were three methods generally used to solve this part.
Method 1. d 2π R
=
t
T υ= For an expression for υ in terms of the period T
For substitution of 2π R for the length d of the orbital path
Solving for T gives T = 2π R υ For correct substitution for υ from (a) i. T= 2π R
4π R
=
GM J
GM J
R
2 1 point
1 point 1 point 3 Method 2. T= 2π ω = 2π R υ For the equation for T in terms of ω
For substitution of υ R for ω in the equation
For correct substitution for υ from (a) i. T= 1 point
1 point
1 point 2π R
4π 2 R3
=
GM J
GM J
R Method 3.
2 points 2 GmM J
mυ 2
æ 2π ö
= mω 2 R = m ç
F=
÷ R=
R
R2
èTø
4π 2 mR GmM J
=
T2
R2
4π 2 R3
2
For T =
GM J 1 point 4π 2 R3
Note: Direct use of T =
was awarded 1 point only, if it was defined as
GM J
2 Kepler’s law or Law of Orbits.
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2001 SCORING GUIDELINES
Question 2 (cont.)
2. (b) 3 points Distribution
of Points For use of correct equation for T from (a) ii. or derivation of this relationship T= 4π R
GM J
2 For correct solution for R or R 3 , numerical or symbolic, from above equation R3 = 1 point 3 GM J T 2
4π 2 1 point 1/ 3 or æ GM J T 2 ö
R=ç
÷
2
è 4π
ø or R = é( 6.67 × 10−11 N • m 2 kg 2 )(1.9 × 10 27 kg )( 3.55 × 104 s )
ê
ë For a correct answer 2 4π 2 ù
ú
û 1/ 3 1 point R = 1.59 × 108 m
Note: If RJ was subtracted from R the answer point was only awarded if the difference
was clearly indicated to be the height of the orbit above the surface.
2. (c) i. 3 points For stating that the orbit is an ellipse
For diagram with orbit drawn completely outside the circle with point of contact only at
point P and major axis along PJ.
Partial credit of 1 point awarded for any path or orbit completely outside the circle.
No points were awarded in any part of path or orbit was inside the circle. 1 point
2 points Copyright © 2001 by College Entrance Examination Board. All rights reserved.
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2001 SCORING GUIDELINES
Question 2 (cont.)
2. (c) ii. 3 points Distribution
of Points
For stating that the orbit is an ellipse
For diagram with orbit drawn completely inside the circle with point of contact only at
point P and major axis along PJ.
Partial credit of 1 point awarded for any path or orbit completely inside the circle.
No points were awarded if any part of path or orbit was outside the circle. 1 point
2 points Note: Three points may also be awarded in this part for a path in which the satellite
“crashes” into Jupiter only if there is specific reference to the scale of the orbit from
part (b) and the given radius of Jupiter. Copyright © 2001 by College Entrance Examination Board. All rights reserved.
Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 9 AP® PHYSICS C: MECHANICS
2001 SCORING GUIDELINES
Question 3
15 points total
3. (a) 3 points Distribution
of Points
1 point For a correct formula for the rotational inertia I = å mr 2 For a sum containing a term of the form mL2 (may include extra incorrect terms, but
the point was not awarded if the expression does not contain an mL2 term) I = mL2 + mL2 For the correct answer 1 point 1 point I = 2 mL 2 3. (b) 6 points For a correct expression of Newton’s 2nd law 1 point For correct substitutions into Newton’s law 1 point For a correct formula for torque
τ = Iα or Tr
Iα = Tr
Iα 1 point F = ma 4mg − T = 4ma T= r From Newton’s 2nd law equation above: T = 4mg − 4ma Substituting into the torque equation:
Iα = 4mg − 4ma r For substituting the expression for I from part (a) into Newton’s law
2mL2α 1 point = 4mg − 4ma r For the expression α = a r
Substituting this expression into the previous equation: 1 point 2mL2 a
= 4mg − 4ma
r2
For the correct answer a= 1 point 2 2 gr
L + 2r 2
2 Note: For the solution a = 4mg − T
, obtained by solving 4mg − T = 4ma for a directly,
4m a maximum of 3 points was awarded for part (b) as follows; 1 point for Newton’s
law, 1 point for substitutions, and 1 point for answer.
Copyright © 2001 by College Entrance Examination Board. All rights reserved.
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2001 SCORING GUIDELINES
Question 3 (cont.)
3. (c) 3 points Distribution
of Points For correctly checking the space in front of “Equal to 4mgD”
For correct justification, such as “The kinetic energy gained by the two smaller blocks
comes from the decrease in the potential energy of the 4m block.” OR “Total energy
is conserved.” 1 point
2 points Note: No points awarded for part (c) if wrong box was checked.
3. (d) 3 points For correctly checking the space in front of “Less”
For correct justification, such as “The small blocks rise and gain potential energy. The
total energy available is still 4mgD. Therefore the kinetic energy must be less than
in part (c).” 1 point
2 points Note: No points awarded for part (d) if wrong box was checked. Copyright © 2001 by College Entrance Examination Board. All rights reserved.
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 Physics, mechanics, Energy, College Entrance Examination Board, college entrance examination, Entrance Examination Board

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